Physics 212 – Midterm Exam I NAME:__Albert Einstein_______

Summer 2003 STUDENT ID#:___007_________

Prof. Milos Steinhart

INSTRUCTIONS: This is a closed book exam. You may use only the one page of formulas provided to you. You must show all of your work in the space provided for each question and you must include appropriate units and significant figures. If you need more space, use the other side of the paper but mark clearly to which problem and which question is your work related. Do not use other papers!

You are bound by the University’s Honor Code. Any evidence of cheating must be reported to the College of Arts and Sciences.

PLEASE CIRCLE ALL ANSWERS

All problems are worth 20 points. Partial credits are shown in brackets.

SCORE

Problem 1 _____20________________

Problem 2 _____20________________

Problem 3 _____20________________

Problem 4 _____20________________

Problem 5 _____20________________

TOTAL ____100________________

Problem 1: A spherical drop of water carrying a charge 30 pC has a potential of 500 V at its surface.

a) (5) What is the radius of the drop?

j = kQ/r Þ r = kQ/j = 9 10⁹ Vm/C 3 10^-11 C /5 10² V = 27 10^-4 /5 =

r = 0.54 mm

b) (5) What is the intensity of the electric field near to the surface of this drop?

Gauss: E = s/e₀ = kQ/r² = j /r Þ E = 500 / 5.4 10^-4 = 9.3 10⁵ V/m

c) (5) If eight such drops of the same charge and radius combine to form a big single spherical drop. What is the potential at its surface?

V₈ = 8 V Þ r₈ = 2 r ; Q₈ = 8 Q Þ j₈ = kQ₈/r₈ = k8Q/2r = 4 j = 2000 V

d) (5) What is the intensity of the electric field near to the surface of the big drop?

Similar as b) Þ E₈ = j₈ / r₈ = 4j/ 2r =2 E = 1.86 10⁶ V/m

Problem 2: A slab of copper of thickness b is inserted into a parallel plate capacitor with the plates of the area A, separated by the distance d (of air). The capacitor is charged but the power source is disconnected before the insertion of the slab.

a) (6) What is the ratio of the capacitance before to that after the slab is introduced?

Original capacitance C₀ = e₀A/d . Since the slab is conductive, the field inside it is zero and the potential in it and on the surface is constant. This means that the gap was effectively reduced to form d to d – b Þ

C = e₀A/(d-b) Þ C₀/C = (d - b)/d

b) (6) Find the ratio of the stored energy before to that after the slab is inserted?

Since the power source is disconnected the charge is conserved Q₀ = Q ;

Energy before insertion: U₀ = Q₀²/2C₀ and energy after: U = Q²/2C ;

U₀/ U = C/C₀ = d / (d - b) Þ U < U₀

c) (6) How much work is done on the slab (by an external agent) as the slab is inserted?

W = U - U₀ = Q²/2 (1/C – 1/C₀) = Q²/2C₀ ((d-b-d)/d) = -bQ²/2C₀d =

= -bQ²/2e₀A

Since W < 0 the field does positive work and the external agent does negative work.

d) (2) Is the slab pulled in by the field or do we have to push it in?

Since U < U₀ and field does work the slab will be pulled between the plates.

Problem 3: At a certain time a charged particle with the mass of m = 20 ng (20 10^-12 kg) was detected in a point a of a uniform electrostatic field. The potential in the point a is j_a = 10⁵ V and the speed of the particle is v_a = 10⁵ m/s. Later the same particle was detected in a point b exactly 1 m to the right from the point a. The potential in the point b is j_b = 8 10⁴ V and the speed of the particle is v_b = 8 10⁴ m/s. The influence of gravitation can be neglected!

a) Describe the electrostatic field.

The field lines are horizontal and point to the right i.e. from A to B.

E = +2 10⁴ V/m

b) What is the work done by the electric field on the particle?

There is only the field and the particle. The work done by the field on the particle is equal to the change of its kinetic energy.

W = E_kb – E_ka = m/2 (v_b² – v_a²) = 2 10^-11 / 2 (-64+100) 10⁸ = -36 10^-3 J

Accelerated movement approach:

v₂ = v₁ + at; s = v₁t + at²/2 Þ t =(v₂ – v₁)/a Þ a = (v₂² – v₁²)/2s = - 1.8 10⁹ ms^-2

F = ma = -36 10^-3 N; W = Fs = -36 10^-3 J

c) What is the charge of the particle?

W = F s = E q s Þ q = W/Es = -36 10^-3 Nm / 2 10⁴ Nm/C = -18 10^-7 C

Since the charge is slowing down its charge must be indeed negative!

q = -1.8 10^-6 C.

q = F/E –1.8 10^-6 C

Problem 4: We have a circuit:

a) What is the topology of the circuit?

2 junctions, 3 branches, 3 loops, 2 loops independent.

b) What is the current? In the point A?

R₂₃=R₂R₃/(R₂+R₃) = 1 W

Let’s expect clockwise direction starting from the point A.

KL: -V₁ + R₁ I + R₂₃ I + V₂ + R₄ I + R ₅ I + R₆ I = 0 Þ

I = (V1-V2)/(R₁+R₂₃+R₄+R₅+R₆) = (24 – 6)/9 = 2 A

Superposition: I₁ (produced by V₁) = 8/3 A, I₂ = - 2/3 Þ I = 2 A

c) Find the potential in the points A, B, C, D with respect of the potential in the point E.

We can ground the point E

j(D) = j(D) + the potential drop on R₄ Þ j(D) = R₄ I = 4 V

Similarly:

j(C) = j(D) + R₃ I + V₂ = 4 + 2 + 6 = 12 V

j(B) = j(C) + R₂ I = 12 + 2 = 14 V

j(A) = j(B) + R₁I – V₁ = 14 + 8 - 24 = -2 V

or j(A) = j(E) – R₅ I = -2 V

d) How the current changes, if we change the V₁ from 24 V to 15 V?

Similarly as a) I = (15-6)/9 = 1 A

Superposition: I_1d = 5/3 A, I_2d = I₂ = -2/3 A Þ I_d = 1 A

e) How the current changes, if we change V₁ back to 24 V and increase V₂ four times?

Now both voltages are the same but opposite polarity. From the superposition principle I = 0!

Or : I_1e = 8/3 A, I_2e = 4I₂ = -8/3 A Þ I_e = 0 A

Problem 5: Multiple choice. Pick the single best answer to each conceptual question below. Please include a brief explanation for partial credit. Each answer is worth 4 pts.

1. An electric dipole in a uniform electrostatic field experiences

a) Only a net external force.

b) Only a torque. (Shift forces cancel)

c) Both a net external force and a torque.

d) Neither net external force nor a torque.

e) Answer depends on the strength of the field.

2. A spherical metal shell in equilibrium carries a uniform positive surface charge. The potential is the same over the surface of the shell. Which statement is correct?

a) The potential is the highest in the geometrical center of the shell.

b) The potential is the lowest in the geometrical center of the shell.

c) The potential at any point within the shell volume is the same as on the shell surface. (Otherwise there would not be equilibrium. Charges would move pushed by the potential and thereby their potential energy difference)

3. Compared with the applied electric field, the electric field within a linear dielectric is

a) Smaller. (K or e_r times)

b) Larger.

c) Depends on the dielectric.

d) The same.

4. Two identical capacitors are connected first in parallel and then in series. Which combination has the greatest capacitance?

a) The pair in parallel. (Since it is a sum of the capacitances)

b) The pair in series.

c) The two combinations have the same capacitance.

5. Two identical resistors are connected first in parallel and then in series. Which combination has the greatest resistance?

a) The pair in parallel.

b) The pair in series. (Since it is a sum of the resistances)

c) The two combinations have the same resistance.